Exercise 1.6

Question

A ring $A$ is such that every ideal not contained in the nilradical contains a non-zero idempotent (that is, an element $e$ such that $e^2 = e 
e 0$). Prove that the nilradical and Jacobson radical of $A$ are equal.

Amit Awasthi · Accepted Answer

\begin{proof}
  $\mathfrak{N} \subseteq \mathfrak{R}$ as in Exercise 4. Conversely, suppose there exists a non-zero idempotent $e \in \mathfrak{R} \setminus \mathfrak{N}$. Then, $e(1-e) = e - e^2 = e-e = 0$. By Prop.~$1.9$, $1-e$ is a unit. But then, $0 = 0(1-e)^{-1} = e(1-e)(1-e)^{-1} = e$, which is a contradiction.
\end{proof}

Exercise 1.6

Answers

Comments

Exercise 1.6

Answers

Comments

Add answer